se properties of limits and algebraic methods to find the limit, if it exists. (If the limit is infinite, enter '[infinity]' or '-[infinity]', as appropriate. If the limit does not otherwise exist, enter DNE.) lim x→−1 f(x), where f(x) = x2 + 5 x if x ≤ −1 4 x 3 − x − 1 if x > −1 lim x→−1+ f(x) = Incorrect: Your answer is incorrect. lim x→−1− f(x) = lim x→−1 f(x) =

Answer:We have the function [tex]f(x)=x^2+5x\; if \;x\leq -1[/tex] and [tex]f(x)=4x^3-x-1\;if\; x>-1[/tex]Then [tex]lim_{x\rightarrow -1}f(x)[/tex] exist if there is the limit of f (x) approaching with values on the right of -1, [tex]lim_{x\rightarrow -1^{+}}f(x)[/tex], the limit of f (x) approaching with left values of f (x), [tex]lim_{x\rightarrow -1^{-}}f(x)[/tex], and both coincide.Observe that the limit of f (x) approaching with values on the right of -1 is [tex]lim_{x\rightarrow -1^{+}}f(x)=lim_{x\rightarrow -1^{+}}4x^3-x-1=4(-1)^3-(-1)-1=-4+1-1=-4[/tex]and the limit of f (x) approaching with values on the left of -1 is [tex]lim_{x\rightarrow -1^{-}}f(x)=lim_{x\rightarrow -1^{-}}(x^2+5x)=(-1)^2+5(-1)=1-5=-4[/tex]Since the two limits coincide, then[tex]lim_{x\rightarrow-1}f(x)=-4[/tex]